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This question came up because of this diagram that I saw in my textbook of an angular simple harmonic oscillator. I've always struggled a bit with torque and rotational dynamics in general, and I thought that I understood it once I saw the relevant equations derived from Newton's laws and the conservation of energy. However, those derivations relied on the fact that there was a force $F$ being applied to the rotating object at distance $r$ from the axis of rotation and that $F \times r \neq 0$. In the case of a force applied at the axis of rotation, however, $r = 0$, and thus $F \times r = 0$ also. How can there be a torque in the case of an angular simple harmonic oscillator, and where is the force coming from that is causing this torque?

This is the webpage with the full material that is covered in my textbook.

Edit: another example where this seems to be happening in my textbook is during a section on the precession of a gyroscope. The book seemed to be saying that the gyroscope experiences torque after it has been set spinning, but that this torque is not applied at the edges of the gyroscope. Rather, it comes from the rod about which the gyroscope frame is rotating. Could the answer to my question in both of these cases be that, in fact, $r \neq 0$? Is $r$ then the radius of the string (in the case of the angular simple harmonic oscillator) and the radius of the central rod (in the case of the spinning gyroscope)? Or is the gyroscope example fundamentally different?

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  • $\begingroup$ Hi forgive my ignorance here, but I am guessing it's inherent in the material that the spring is composed of and the mass of the disc at the bottom is just accentuating a normal restorative force. Sorry if I misunderstood your question regards $\endgroup$ – user74893 Apr 13 '15 at 20:57
  • $\begingroup$ Just to clarify, by spring I mean the wire, cord, string whatever, holding the bottom mass , that has by its composition , say of separate wire stands, a natural tendency to unwind, providing the turning force. $\endgroup$ – user74893 Apr 13 '15 at 21:05
  • $\begingroup$ So could my confusion be coming from the assumption that the wire is infinitely thin? Would the equations be useless in such an idealized scenario with an infinitesimally thin massless wire? $\endgroup$ – jlftfeisall Apr 13 '15 at 21:11
  • $\begingroup$ I hope you get a proper answer below, but almost any thing , say a common vacuum cleaner hose, for example, will resist your attempts to twist it by holding one part still and using your other hand to rotate it, in other words it's a spring, although I admit it does not look like one. Hope this makes sense. A spring is not always an obviously coiled up looking thing. $\endgroup$ – user74893 Apr 13 '15 at 21:17
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Yes, you are correct that in the case of the oscillating disk, the suspending string is not acting exactly at $r=0$. The attachment to disk is spread over a finite distance and is capable of supplying off-axis forces. If you hypothesize some sort of infinitesimally thin wire, you then have the problem of what it means for such an object to twist or rotate.

The gyroscope does not depend on the radius of the central rod. Instead the $r$ in this case is the distance from the center of mass that the force is applied (that distance would be $L \sin (\theta)$ in the diagram).

In your gyroscope diagram, the table mount is pushing up on the rod. But this force does not pass through the center of mass of the gyroscope and produces a torque.

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