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Consider a toilet paper roll floating ISS. Can an astronaut unfurl the roll only by pulling the loose end of paper (and not holding the rest of the roll in place, for example by using his finger as an axle)? I believe the answer is no, since we cannot create the torque needed to accomplish this unless the roll is fixed in place. What I suspect will happen is that the roll will not undergo any rotational motion, only translational motion.

Probably this question is trivial for those who know about rotational equilibrium, but I have not yet learned this.

math_lover
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    Since the astronaut is not pulling along the line trough the center of mass of the roll, there will be a torque, which will lead to a rotation. After the first pull the roll will unroll itself. – CuriousOne Jan 04 '15 at 00:07
  • what will the torque come from if the astronaut is weightless too ? how can you say the astronaut won't just pull himself toward the toilet roll i suspect it will have a little bit of an effect on everything and since that's all it would take to unroll the entire roll i would have to agree with CuriousOne, i like the question though :) – Daz Hawley Jan 04 '15 at 00:20
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    It will unroll and drift towards the puller – Phoenix87 Jan 04 '15 at 00:23
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    The question is of great practical importance for the propulsion of the spacecraft... how do we have to angle an engine that is displaced relative to the intended trajectory going trough the center of mass to avoid a steady torque turning the spacecraft off course? – CuriousOne Jan 04 '15 at 00:32
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    Comrade @CuriousOne, you are not authorized to tell the world about the toilet-paper-powered rocket project. Please report to re-education centre 3 at your earliest convenience. – David Richerby Jan 04 '15 at 12:03
  • @CuriousOne Sounds like angular momentum is not preserved in that case. So something must be missing from that explanation. – kasperd Jan 04 '15 at 13:21
  • @kasperd: Why would angular momentum not be conserved? The angular momentum that you are missing is in the exhaust plume that is also off-center. – CuriousOne Jan 04 '15 at 15:31
  • @CuriousOne I was referring to your first comment in which you suggested that one object would start rotating, by being pulled off-center. But the force on the one doing the pulling would not have to be off-center. So it appears that scenario does not preserve angular momentum since one object starts rotating while the other does not. – kasperd Jan 04 '15 at 15:51
  • @kasperd: The astronaut object will be moving slightly perpendicular to the pull in the center of mass system. You could petition NASA to send a roll to the ISS and have them demonstrate. – CuriousOne Jan 04 '15 at 15:55
  • @CuriousOne If the astronaut started out not moving at all, then the movement caused by the pulling force would have to be parallel to that force. However thinking about the center of mass did make me realize what would be happening. When two objects orbit a comment center of mass then the orbiting introduces an angular momentum. And forces can transfer angular momentum between the orbiting and the rotation of the individual objects. – kasperd Jan 04 '15 at 16:15
  • @kasperd: That sounds like a good way of thinking about it. – CuriousOne Jan 04 '15 at 16:19
  • If the toilet paper has 0 mass then I think no torque would be generated. But with mass, it has inertia, and will thus be subject to rotational inertia rules. – Buttle Butkus Feb 27 '15 at 02:48

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The roll doesn't have to be in a fixed place for a torque to be generated. Torque will be generated regardless of a fixed axis or not and your toilet paper will unroll eventually. Torque is generated because as @CuriousOne said above, the force is acting somewhere other than through the centre of mass thus creating an unbalanced force on one side which creates a rotation about that centre of mass.

If the roll is not fixed on something, then most probably the initial pull will create some linear motion as well as a rotation so the roll will unroll itself while drifting in space.

PhotonBoom
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  • And the astronaut will also be pulled toward the roll, with their torque dependent on how they pull relative to their own center of mass. Of course since they have more mass they will acquire less velocity (F=mv and so on) so this may or may not be noticable. – keshlam Jan 04 '15 at 02:23
  • This is very interesting. I would like to see the astronauts actually do this as a demo. A similar question, which is probably also trivial: If you have a rod which is initially parallel to the ground, with ball masses at each end of the rod - a heavy mass and a ligth mass- and you drop the rod, will it rotate as it falls? I suspect the answer is no because the torque due to gravity cancels out (how do we prove this?). @keshlam: F=mv? Hmm I don't think thats correct! :) – math_lover Jan 04 '15 at 03:43
  • @JoshuaBenabou, since the distance of the two balls from the centre of mass of the rod is the same, and they both experience an qual acceleration $g$, the mass with bigger mass will experience a bigger force thus creating an unbalanced net torque. – PhotonBoom Jan 04 '15 at 03:51
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    @JoshuaBenabou: Tupogharpcal eror, sorry. F=ma, of course. – keshlam Jan 04 '15 at 04:00
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    (BTW, can I volunteer one of my cats for the unroll-toilet-paper-in-space experiment? She's proven to be an expert at doing it on Earth, no matter what barriers I put in the way.) – keshlam Jan 04 '15 at 04:54
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    You can conduct the experiment on earth, just put the toilet roll on a smooth table and start pulling, you'll see it start rotating and start translating. – QCD_IS_GOOD Jan 04 '15 at 05:33
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    @PhotonicBoom, I actually performed the experiment with an unbalanced rod, but it remained essentially level throughout the fall. Perhaps the fall time was too short to notice any rotation. – math_lover Jan 04 '15 at 06:02
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    @JoshuaBenabou assuming your rod falls through air and the two balls have a different relative air resistance (e.g. air balloon versus massive stone), it will rotate due to the drag. The gravitation alone will not do this. – Paŭlo Ebermann Jan 04 '15 at 16:20
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    @PaŭloEbermann that's absolutely correct for a free falling rod. – PhotonBoom Jan 04 '15 at 16:24
  • @keshlam — F = mv ? You mean F = ma, don't you ? – Nicolas Barbulesco Jan 04 '15 at 17:20
  • @PhotonicBoom: so will the rod rotate or not (assuming no air resistance)? – math_lover Jan 04 '15 at 17:58
  • @JoshuaBenabou, is it free-falling? If it is then there will be no rotation due to gravity as everything (even the centre of mass) is now accelerating downwards at the same rate. Also bear in mind that the centre of mass of the whole system is now located closer to the heavy mass, so the distances of the two masses from the centre of mass are now different. – PhotonBoom Jan 04 '15 at 18:16