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Imagine that you are standing (naked) in the center of a level circular disk (with a radius of, say, 3 meters) that is completely frictionless.

Without resorting to blowing air or excreting any other sort of air (or fluids), would one be able to reach the edge of the disk?


Additional assumptions (from the comments):

  • you are not allowed to excrete any solids either...
  • let's assume that you are in a vacuum
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  • $\begingroup$ Using liquids or gases is forbidden, but what about solids? $\endgroup$ – DBPriGuy Dec 15 '16 at 22:38
  • $\begingroup$ @DBPriGuy Good point, let's assume that nothing is allowed to come out of (or off of, as in somehow removing a body part) your body. $\endgroup$ – Erik M Dec 15 '16 at 22:40
  • $\begingroup$ @ErikM: What about moving air with hands? $\endgroup$ – akhmeteli Dec 15 '16 at 22:42
  • $\begingroup$ I don't know about the reputation of this, but this may be relevant: science20.com/hammock_physicist/swimming_through_empty_space $\endgroup$ – DBPriGuy Dec 15 '16 at 22:42
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    $\begingroup$ So you are naked, in a vacuum and not allowed to expel anything - nor (I suppose) call for help. Are you trying to find a solution or are you trying to find a situation which you cannot possibly get out of? $\endgroup$ – sammy gerbil Dec 16 '16 at 3:29
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Expanding on the suggestion made by @LastStar007 :

I assume that you and the disk are floating in space with a very weak gravitational attraction.

You are standing above the centre of the disk so jumping up at this point will not get you to the edge of the disk. Relative to you, the disk will move away, remaining parallel to its original direction. You need to create an impulse which also causes the disk to turn.

So first slide your feet apart, crouch down to flex your muscles, then jump pushing with one foot only. The result should be that the disk rotates about its CM while the CM also moves away from you. If you apply the appropriate impulse then the disk will make approx. $\frac14$ rotation in the time that its CM moves one radius away from you. Then the disk will collide with you at its edge.


The required impulse depends on the mass, radius and depth of the disk, your mass and the separation of your feet.

enter image description here

Suppose that the disk has width $2R$ and depth $2a$ and that your feet are $2r$ apart when you jump with one foot. You and the disk have masses $m$ and $M$. Apart from being able to spread your feet wide I shall assume that you have negligible size and that your CM is initially close to the surface of the disk.

Your jump applies impulse $I$ which causes your CM to move away with speed $v_1$ and the CM of the disk to move away with speed $v_2$, where
$I=mv_1=Mv_2$.
The disk acquires angular velocity $\omega_2=\frac{rI}{J_y}$ where $J_y=\frac{M}{12}(3R^2+a^2)$ is the moment of inertia of the disk (=cylinder) about an axis through a central diameter. You also acquire a rotation but I shall ignore this. If you wished you could adapt the calculation to take your finite size into account. If your dimensions are comparable with the disk it will affect the angle through which the disk has rotated when it collides with you.

In the right-hand diagram the CM of the disk moves from P to X in the same time $t$ that you move from distance $a$ above P to Y. So
$t=\frac{PX}{v_2}=\frac{PY-a}{v_1}$.
Meanwhile the disk has turned through angle $\omega_2 t=\frac12\pi-A$ where $\tan A=\frac{a}{R}$. We also have $PX+PY=XY$ where $XY^2=R^2+a^2$.

Combining all of the equations might not lead to an easy formula for the impulse $I=mv_1$; a numerical solution might be necessary, especially if your finite size is taken into account.

If the disk is very thin $(a \ll R)$ then angle $A=0$ and distance $XY=R$. Reaching the edge of the disk does not depend on the size of the impulse, only its distance $r$ from the CM of the disk :
$\frac{r}{R}=\frac18\pi(\frac{M}{m}+1)$.
Depending on the relative masses of you and the disk, and the distance you can spread your feet compared with the diameter of the disk, it may not be possible to spread your feet wide enough to reach the edge of the disk when you jump up.

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In principle, you can use radiation pressure of sunlight (or other external source of light).

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A paper published in 2003 demonstrated that in a gravity well* (assuming GR holds):

...translation in space can be accomplished simply by cyclic changes in the shape of a body, without any thrust or external forces.

Thus, translation can be achieved in a vacuum through cyclic body motion changes. So, our poor victim willing test subject can escape this frictionless death trap interesting experiment with just the right movements.


* I specify a gravity well just to emphasize that this only works in curved space-time.

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You jump. The disk still gravitates.

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  • $\begingroup$ I'm not sure if I follow. The disk would move (horizontally) relative to the person if they jumped? $\endgroup$ – Erik M Dec 15 '16 at 22:51
  • $\begingroup$ @Erik M At the time, your question said get off the surface. $\endgroup$ – LastStar007 Dec 15 '16 at 22:52
  • $\begingroup$ Yes, I'm sorry, I realized that was ambiguous (apologies for changing the meaning). $\endgroup$ – Erik M Dec 15 '16 at 22:55
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    $\begingroup$ The question at the time of posting prompted no critique or clarification, and my answer answered that question adequately. Am I supposed to keep up with the question's revisions and constantly revise my answer to reflect them? $\endgroup$ – LastStar007 Dec 19 '16 at 5:52
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If you move a hand, momentum conservation will slide you to the edge, (no friction you said). The momentum comes from the chemical reactions inducing the motion in the hand, ultimately radiated electromagnetic energy.

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    $\begingroup$ I think the physics you are using is flawed. The motion of your hand would cause your body to move in the opposite direction consistent with conservation of linear momentum, but as soon as your arm is fully extended, your hand will decelerate, and both your body and your hand will come to rest once again relative to each other and the disk, making this tactic unusable for escape. $\endgroup$ – UniqueWorldline Dec 18 '16 at 18:40
  • $\begingroup$ You also mention chemical reactions and EM energy. The only possible explanation I can conceive of that you might be alluding to is radiation pressure off the surface of your skin. The problem with this hypothesis is that this emission of radiation is essentially isotropic. Different regions of the body might give off a greater flux of EM (almost all infrared) radiation, but it would still be emitted essentially evenly in all directions, so there would be no net force acting on you. $\endgroup$ – UniqueWorldline Dec 18 '16 at 18:44
  • $\begingroup$ @UniqueWorldline Once set in motion since the plane is frictionless you will end up over the edge of the disk. There is aburst of chemical reaction when a motion is set up, and the radiation it has to be directional , infrared photons leaving in a bunch. I accept the objection that most of momentum conservation will be taken up by the rest of the body in the opposite direction, but even a small amount of directional radiation since there is no friction should be sufficient to slide of the disk. $\endgroup$ – anna v Dec 18 '16 at 19:34

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