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Just for example: assume an iron bar one foot in length. If you push on one end, the entire bar will move. This seems instantaneous. but actually, from my understanding, the atoms all push against each other in a very fast "wave" - making the entire bar move.

Now, say the bar is 2 light-years long. We are at one end of the bar, and rotate it 90 degrees. Clarification: rotation is the same as if you held a pencil horizontally by the eraser, then turned it vertical. This seems to mean the other end of the bar stays where it started until our motion travels through the entire bar. How long would that take?

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  • $\begingroup$ So you mean rotate it about its long axis? $\endgroup$
    – jpmc26
    Jan 9 '15 at 23:10
  • $\begingroup$ Although not a duplicate I guess, the answers to this question are the same as the answers to this one: physics.stackexchange.com/questions/2175/… $\endgroup$
    – tpg2114
    Jan 10 '15 at 4:47
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This is actually a really interesting question, but the answer is simpler than you might think. The pressure wave propagates along the bar at the speed of sound in whatever material the bar is made of. This is because sound is nothing more than pressure waves, and so the speed of sound is by definition exactly what you're looking for.

For example, the speed of sound in iron is around 5130 m/s (from this site), so any disturbances would propagate along the bar at that speed.

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    $\begingroup$ With only considering these factors, it would seem to take 116,878 years (with some rounding) for the opposite end to start moving. If anyone else was curious. $\endgroup$ Jan 9 '15 at 17:48
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    $\begingroup$ How does this slow speed of propagation affect how much mass the object appears to have? Or is this best answered by a new question? $\endgroup$
    – Michael
    Jan 9 '15 at 21:46
  • $\begingroup$ What do you mean by how much mass the object "appears" to have? $\endgroup$ Jan 10 '15 at 0:21
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    $\begingroup$ @JedThompson I think he might be asking what happens if you yank on one end, in the time before the movement propogates to the other end. Since only so much of the object has "seen" the energy you applied to your end of it, one might think that the observed mass from monitoring the amount of force applied to the bar might be lower than the actual mass of the bar. If that makes any sense. $\endgroup$
    – Schilcote
    Jan 10 '15 at 3:58
  • $\begingroup$ @Schilcote: Ok, that makes sense, but I think the better way to understand that thought experiment is to just recognize that it's a really bad way to measure the mass of the bar. If you tried to hook up a force probe to the bar and measure the force you are applying and how far the bar has moved, I think you'd basically end up measuring how hard it is to compress whatever material the bar is made of (until the pressure wave reaches the other end that is). I'm not 100% sure about this however, but I am sure that this is not a very good way to measure the mass of the bar. $\endgroup$ Jan 10 '15 at 6:28
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You can grab on end and pull it or push it, but since you want to rotate it, and you have to travel a quarter circle of a 1 light year radius circle, it's going to take you more than $2\pi /4$ years to get your end into position.

And as you grab your end each part of the rest of the bar stays at rest until the pressure is transmitted from your hand through the bar to the part in question, and pressure waves travel at the speed of sound since sound is a pressure wave. You can choose to move at less than the speed of sound in the bar yourself to give it a chance to try to move somewhat together. If you move your end faster then bar must and will deform, so you can imagine that you have to pull quite hard. The faster you try to move it, the more it will deform and the harder it gets. Deform it too much and it can even break.

And if you try to stop turning it after the 90 degree turn that stop will also take time to propagate. And when you deform it it can start to wobble, it's a bit like moving a slinkly or a rolled out piece of playdough, you might even lose some energy to heat as the speed of sound itself can vary as the objects deforms enough to change it's density. So you might end up with a vibrating bar despite your best efforts.

Another thing to consider about a bar that large is that it might self gravitate as well.

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    $\begingroup$ @DoubleDouble It will start deforming, and that deformation can propagate at the speed of sound in the material, but if you are going faster than sound it's not going to keep up and the deformation near you gets worse even though more of the bar is deforming. When something deforms too much it can break. When it breaks the other half will snap back, and all those pressures keep moving down that bar. $\endgroup$
    – Timaeus
    Jan 9 '15 at 17:53
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    $\begingroup$ So long as the deformation does not exceed the elastic limit, the bar will bend without breaking and without permanently deforming. Steel, when properly tempered, is quite elastic. $\endgroup$
    – Hot Licks
    Jan 9 '15 at 18:03
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    $\begingroup$ @DoubleDouble A twist around the long axis would propagate more slowly than a push parallel to the long axis, because a twist produces an S-wave whereas a push produces a P-wave. $\endgroup$
    – zwol
    Jan 9 '15 at 19:46
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    $\begingroup$ The end result of all this seems to be - I would also need to calculate the duration for my intended rotation such that the deformation will not exceed the elastic limit of the object - or else it will permanently bend or break. While this has no effect on how long it will take the opposing side to start moving, it certainly has a great effect on how long the entire object would take to finish rotating. $\endgroup$ Jan 9 '15 at 20:10
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    $\begingroup$ @MooingDuck - Eventually (assuming there was some damping present) the motion would become indistinguishable from Brownian motion. At the very least there would be damping inherent in the material, but steel has very little. The bar could "ring" for a long time. $\endgroup$
    – Hot Licks
    Jan 9 '15 at 23:15

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