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I haven't studied much about this, so if I am mistaken about something please correct me.

From what I have seen around the Internet, a force applied to a object takes time to propagate through the object because it has to interact with the molecules around the ones that the force was applied to. That means that for the force to get to the other side of the object it will take time especially if it is over a large distance. To my understanding the displacement of those molecules propagating through the rest of the object can be called a deformation wave. This deformation wave travels at the speed of sound through the medium (the object).

What if the object that the force was being applied to was a wheel with a bar from the top of the wheel to the bottom that crosses the center? The bar is made out of the same material as the wheel and is bonded to the wheel (there are no gaps between the bar and the wheel). If a force was applied to the top of the wheel would the deformation wave travel half the circumference of the wheel or the diameter bar to get to the other side? Basically what I am asking is:

  1. Is the information presented in this question accurate, or am I mistaken about someting?
  2. Does the deformation wave move at the speed of sound through the medium, or is it even necessary for the answer to this question?
  3. And does the deformation wave find the fastest path to the opposite side of the wheel (the bar)? Does it go in all directions? Or would it just go the direction that the force was applied to (the outside of the wheel)?
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  • $\begingroup$ BTW look up stress waves. $\endgroup$ – ja72 Apr 16 '14 at 23:39
  • $\begingroup$ It probably is worth mentioning that while the sudden exertion of a force on a semi-rigid object will cause propagating sound/force waves, these waves are transient, and in the long term, the steady-state force that moves the wheel can basically be modeled as a regular rigid-body-type force. So the question is mostly relevant for the first couple milliseconds of the object's dynamics when a force is exerted on it. $\endgroup$ – DumpsterDoofus Apr 17 '14 at 0:08
  • $\begingroup$ However, sometimes it's not accurate to throw away the wave propagation component of forces and only use rigid-body dynamics when modeling things; for example, I recall reading (in some other PhysicsSE question) that a quantitatively accurate simulation of the behavior of Newton's Cradle actually requires one to account for the fact that the forces are largely due to propagating sound waves as the balls impact. So the OP's question actually shows up in some interesting situations. $\endgroup$ – DumpsterDoofus Apr 17 '14 at 0:11
  • $\begingroup$ Please try to simplify the description of the experiment with the wheel; Is it relevant wheter it's rotating? A wheel with spokes/ a disk? Other than that, I think it's an nice question. $\endgroup$ – Volker Siegel Apr 17 '14 at 0:26
  • $\begingroup$ it is not rotating until the force is applied to it, and the only thing that could be considered a spoke is the rod in the middle. I imagined it like how bytbox put it,"just a thin, large torus". $\endgroup$ – Noah Geren Apr 17 '14 at 0:30
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The deformation wave will travel in both directions - there's no way for it to "know" the shortest path. And the resulting set of vibrations will interfere with each other in interesting ways, causing complicated resonances.

So, let's look at a simpler example: just a thin, large torus. We'll look at two points $90°$ apart from one another; one at $\theta = 0$, the other at $\theta = \frac\pi 2$. Quickly apply a sharp force to the point at $\theta=0$, and then let go. If we do this quickly enough, two waves will resulting - one travelling in the negative $\theta$ direction, and the other straight towards the second point. So, sitting at the second point, we would feel one jolt at $t=1$, then another (moving in the opposite direction) at $t=3$, then another at $t=4$, and so on. We've created two waves, and set them moving in a circle.

Figuring out what happens with more complicated geometries is, well, more complicated.

Oh, and a quick note to clear up part 3 of your question: if I take a rod, oriented right-left, and grab the middle, and shove it to the right, both the right side and the left side feel a force. Just $\vec F$ points in a certain direction doesn't mean it only affects things in that direction. So yes, the wave will travel both ways.

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It is true that the deformation wave travels at the speed of sound, but you have to get away from thinking of objects as rigid if you ask about deformation waves. One good image is striking a foam ball with your fist. The overall shape of the ball will change as the ball wraps around your fist. Some of the deformation will travel across the diameter of the ball and will get across in the diameter divided by the speed of sound. Foam is highly damped, so there will not be much oscillation. A second image is blowing on the side of a large soap bubble. Bubbles have low stiffness and low damping, so the wave travels slowly and there is a lot of oscillation. The deformation wave is a local phenomenon, travelling based on what is seen locally until it spreads through the object.

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  • $\begingroup$ The reason I ask this question is because my friend thought that the wave created by the force acting on the wheel would be close to the speed of light. But this could never be true, could it? $\endgroup$ – Noah Geren Apr 17 '14 at 0:02
  • $\begingroup$ No, the highest speed of sound in an element I see is carbon at 16200 m/sec, far from $3\cdot 10^8$ $\endgroup$ – Ross Millikan Apr 17 '14 at 0:05
  • $\begingroup$ "The reason I ask this question is because my friend thought that the wave created by the force acting on the wheel would be close to the speed of light. But this could never be true, could it?" Wow, that's pretty, uh, fast. You're right that it's a lot slower. $\endgroup$ – DumpsterDoofus Apr 17 '14 at 0:18
  • $\begingroup$ This is a nice demonstration video of how a deformation wave is a local phenomenon. $\endgroup$ – fibonatic Apr 17 '14 at 0:36
  • $\begingroup$ Yes that is where I got some of the information for the question, but I was justing wanting to further explore what they had shown. $\endgroup$ – Noah Geren Apr 17 '14 at 0:38

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