1) Would it be correct to state the following: "The longer the radial distance (the distance from the center of the gyro disc) and the lower the height from ground to disc, the greater the angular momentum, and, thus, the more stable the gyro." It seems to me that torque must, somehow, be related to the matter, but not sure if I am on the right track by taking the radial distance as starting point.

2) Are there other differences in the properties of the gyroscope illustrations shown below?

Note: I received the questions exactly as I present them here from the organization I got them from. Therefore, I suffer from not being able to provide more context and specifications. I am only being able to provide my assumptions based on my preliminary knowledge.

Please note that I am a beginner (maybe intermediate) at physics.

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  • $\begingroup$ 3D figures are essential to get a sense of the mass distribution of your gyros, in particular if they have a symmetry axis. $\endgroup$ Sep 3 '18 at 0:54
  • $\begingroup$ @ZeroTheHero - I expect the gyroscopes do have a symmetry axis. The figures are cross sections of the gyroscopes. You get the 3D figure by rotating each figure around the vertical line that crosses the point at the bottom. I would also guess the intent is for the mass of each gyroscope to be the same. $\endgroup$
    – mmesser314
    Sep 3 '18 at 1:10
  • $\begingroup$ Firstly, thanks for your comments. They are, indeed symmetric and their weight is distributed equally. And, yes, their masses are the same too. I assume that that it is not necessary to draw a 3D figure then, am I right? $\endgroup$ Sep 3 '18 at 12:47
  1. A spinning gyroscope has an angular momentum vector $\vec L$ pointing along its axis, given by the right-hand rule. Gravity exerts a torque on the gyroscope, trying to rotate it around an axis perpendicular to its current axis, changing its angular momentum by $d\vec L$ in a time $dt$. Since the axes are perpendicular, $\vec L$ and $d\vec L$ are perpendicular. Thus, the magnitude of the angular momentum does not change, only its direction is changing, causing it to precess. Torque is the rate of change of angular momentum: $$\vec{\tau} = \vec r \times \vec F = \frac{d \vec L}{dt}$$ The ground exerts a normal force, equal to the weight in magnitude, upwards on the point of contact with the ground, while the weight pulls it down at the disk, forming a torque couple given by $\vec r \times \vec F = rF \sin \theta$ above, where $\vec r$ is the vector from the point of contact to the disk and $\theta$ is the angle of the axis from the vertical. The height of the disc above the point of contact with the ground, which is proportional to $r$, affects the lever arm of gravity, causing a larger torque when the gyroscope is "taller" and smaller torque when the gyroscope is "shorter". We are assuming that the point of contact with the ground does not slip/move, and hence using it as our origin.

  2. The size of the disc affects its moment of inertia. Torque is the rate of change of angular momentum with respect to time. The larger and heavier the disc, the greater its moment of inertia and angular momentum and the more resistant it is to changes in its angular momentum and will thus precess more slowly. This is also why the gyroscope precesses faster and faster as it slows down due to friction.

  • $\begingroup$ Thanks a lot, that clarifies a lot! Do I get it correctly that the "F" in this torque formula corresponds to the total weight of the gyroscope and the "r" to the height of the disc? Or does the "r" just stand for the radius of the disc, as would be the case when applying torque calculations on a seesaw, or is it a kind of combi of both factors? $\endgroup$ Sep 3 '18 at 13:30
  • $\begingroup$ @Englishterian: $\vec r$ is the vector from the point of contact with the ground to the center of mass of the gyroscope. I have made my answer more clear. $\endgroup$ Sep 3 '18 at 13:40
  • $\begingroup$ thanks a lot for your praiseworthy answer and the source, I got it now. $\endgroup$ Sep 3 '18 at 15:41
  • $\begingroup$ @Englishterian: You're welcome. $\endgroup$ Sep 3 '18 at 15:42

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