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The reason for this is that centrifugal force $F_{\mathrm{Cent}} = -m \Omega \times \Omega \times r$ is only one of two inertial forces that is introduced when one is in a steadily rotating frame. The other inertial force introduced is the Coriolis force $F_{\mathrm{Cor}} = -2 m \Omega \times v$, which acts on moving bodies in the rotating reference frame in a direction orthogonal to the motion (and to the axis of rotation). Strictly speaking, one has to take into account the effect of both inertial forces when performing Newtonian mechanics in a steadily rotating frame. As it turns out, the Coriolis force has a negligible impact on the dynamics when rotating around the $y$-axis, but dominates the dynamics when rotating around the $x$-axis, which is why the preceding discussion is accurate in the former case but not the latter.

In more detail: suppose we are rotating around the $y$-axis as in the above discussion. In the rotating frame of reference, and starting with a configuration slightly out of equilibrium, we have as before that the $m$-masses experience a little bit of centrifugal force and begin to slide away from the $y$-axis and into the rest of the $yz$-plane. When doing so, they will then experience some Coriolis force in a direction parallel to the $x$-axis (which direction it is depends on the orientation of the rotation, as per the right hand rule formula for the cross product). However, due to the rigidity of the disk, as mediated by tension forces within the disk, it is not possible for the $m$-masses to actually move in the $x$-direction without also moving the much heavier $M$-masses. But the magnitude of the Coriolis force is proportional to the small mass $m$ rather than the large mass $M$, so by Newton's law $F=Ma$ for the $M$-masses, the Coriolis force (or more precisely, the tension force produced in response to the Coriolis force) actually barely affects the motion of the $M$-masses, which stay put on the $x$-axis, and the $m$-masses therefore remain constrained to the $yz$-plane and cannot actually experience any significant motion in the direction of the Coriolis force. (This is what the sentence in the original explanation regarding how tension forces "barely budge" the $M$-masses was referring too, albeit somewhat obliquely.) The analysis of the original explanation now proceeds as before.

In contrast, suppose that the disk is close to the stable equilibrium state when it rotates around the $x$-axis. Working in a rotating frame around this axis, the $M$-masses are near the $x$-axis of rotation, while the $m$-masses lie near the $y$-axis. As before, the $M$-masses experience centrifugal force and thus begin drifting slightly away from the $x$-axis. But then the Coriolis force on these masses kicks in, which is now proportional to the heavy mass $M$ rather than the light mass $m$. As such, the rigid tension forces connecting the $M$-masses to the much lighter $m$-masses offer very little resistance to the Coriolis force, and the motion of the $M$-mass begins to rotate, constantly experiencing Coriolis acceleration in a direction orthogonal to its motion. This effect tends to make the $M$-mass rotate in a tight circle, and basically neutralises the net effect of the centrifugal force by rotating any outward motion away from the $x$-axis back into inward motion. The end result is that in the rotating frame, the disk wobbles a bit around its equilibrium state, but does not dramatically depart from it, which is what one would expect from a stable equilbrium (imagine now a marble rolling frictionlessly around the trough of a valley).

The reason for this is that centrifugal force $F_{\mathrm{Cent}} = -m \Omega \times \Omega \times r$ is only one of two inertial forces that are introduced when one is in a steadily rotating frame. The other inertial force introduced is the Coriolis force $F_{\mathrm{Cor}} = -2 m \Omega \times v$, which acts on moving bodies in the rotating reference frame in a direction orthogonal to the motion (and to the axis of rotation). Strictly speaking, one has to take into account the effect of both inertial forces when performing Newtonian mechanics in a steadily rotating frame. As it turns out, the Coriolis force has a negligible impact on the dynamics when rotating around the $y$-axis, but dominates the dynamics when rotating around the $x$-axis, which is why the preceding discussion is accurate in the former case but not the latter.

In more detail: suppose we are rotating around the $y$-axis as in the above discussion. In the rotating frame of reference, and starting with a configuration slightly out of equilibrium, we have as before that the $m$-masses experience a little bit of centrifugal force and begin to slide away from the $y$-axis and into the rest of the $yz$-plane. When doing so, they will then experience some Coriolis force in a direction parallel to the $x$-axis (which direction it is depends on the orientation of the rotation, as per the right hand rule formula for the cross product). However, due to the rigidity of the disk, as mediated by tension forces within the disk, it is not possible for the $m$-masses to actually move in the $x$-direction without also moving the much heavier $M$-masses. But the magnitude of the Coriolis force is proportional to the small mass $m$ rather than the large mass $M$, so by Newton's law $F=Ma$ for the $M$-masses, the Coriolis force (or more precisely, the tension force produced in response to the Coriolis force) actually barely affects the motion of the $M$-masses, which basically stay put on the $x$-axis, and the $m$-masses therefore remain essentially constrained to the $yz$-plane and cannot actually experience any significant motion in the direction of the Coriolis force. (This is what the sentence in the original explanation regarding how tension forces "barely budge" the $M$-masses was referring too, albeit somewhat obliquely.) The analysis of the original explanation now proceeds as before.

In contrast, suppose that the disk is close to the stable equilibrium state when it rotates around the $x$-axis. Working in a rotating frame around this axis, the $M$-masses are near the $x$-axis of rotation, while the $m$-masses lie near the $y$-axis. As before, the $M$-masses experience centrifugal force and thus begin drifting slightly away from the $x$-axis into the $xz$-plane. But then the Coriolis force on these masses kicks in, which is now proportional to the heavy mass $M$ rather than the light mass $m$. As such, the rigid tension forces connecting the $M$-masses to the much lighter $m$-masses offer very little resistance to the Coriolis force, and the motion of the $M$-mass begins to rotate out of the $xz$-plane, constantly experiencing Coriolis acceleration in a direction orthogonal to its motion. This effect tends to make the $M$-mass rotate in a tight circle, and basically neutralises the net effect of the centrifugal force by rotating any outward motion away from the $x$-axis back into inward motion. The end result is that in the rotating frame, the disk wobbles a bit around its equilibrium state, but does not dramatically depart from it, which is what one would expect from a stable equilbrium (imagine now a marble rolling frictionlessly around the trough of a valley).

One can see this effect qualitatively from Newtonian first principles such as $F=ma$ (as opposed to Hamiltonian or Lagrangian principles, such as that obtained from conservation of energy and angular momentum) by looking at a degenerate case, when one moment of inertia is very small and the other two are very close to each other.

More specifically, consider a thin rigid unit disk, initially oriented in the $xy$ plane and centred at the origin $(0,0,0)$. We make the "spherical cow" hypotheses that this disk has infinitesimal thickness and mass, but infinite rigidity. On this disk, we place heavy point masses of equal mass $M$ at the points $(1,0,0)$ and $(-1,0,0)$ on the $x$ axis, and light point masses of equal mass $m$ at the points $(0,1,0)$ and $(0,-1,0)$ on the $y$ axis. Here $0 < m \ll M$, i.e. m should be viewed as negligible with respect to $M$. (The moments of inertia are then $2m, 2M, 2(m+M)$, though we will not explicitly use these moments in the analysis below.)

UPDATE, September 2019: Due to renewed interest in this question, I will expand upon my 2014 comment regarding why the above analysis also seems to lead to the incorrect conclusion that the disk rotation is also unstable if it one instead rotates around the $x$-axis (so that it is now the $M$-masses that are stationary and the $m$-masses that are rotating), or equivalently if one swaps the location of the $m$-masses and $M$-masses (which we will not do here to try to reduce confusion).

The reason for this is that centrifugal force $F_{\mathrm{Cent}} = -m \Omega \times \Omega \times r$ is only one of two inertial forces that is introduced when one is in a steadily rotating frame. The other inertial force introduced is the Coriolis force $F_{\mathrm{Cor}} = -2 m \Omega \times v$, which acts on moving bodies in the rotating reference frame in a direction orthogonal to the motion (and to the axis of rotation). Strictly speaking, one has to take into account the effect of both inertial forces when performing Newtonian mechanics in a steadily rotating frame. As it turns out, the Coriolis force has a negligible role on the dynamics when rotating around the $y$-axis, but dominates the dynamics when rotating around the $x$-axis, which is why the preceding discussion is accurate in the former case but not the latter.

One can see this effect qualitatively from Newtonian first principles such as $F=ma$ (as opposed to Hamiltonian or Lagrangian principles, such as conservation of energy and angular momentum) by looking at a degenerate case, when one moment of inertia is very small and the other two are very close to each other.

More specifically, consider a thin rigid unit disk, initially oriented in the $xy$ plane and centred at the origin $(0,0,0)$. We make the "spherical cow" hypotheses that this disk has infinitesimal thickness and mass, but infinite rigidity. On this disk, we place heavy point masses of equal mass $M$ at the points $(1,0,0)$ and $(-1,0,0)$ on the $x$ axis, and light point masses of equal mass $m$ at the points $(0,1,0)$ and $(0,-1,0)$ on the $y$ axis. Here $0 < m \ll M$, i.e. $m$ should be viewed as negligible with respect to $M$. (The moments of inertia are then $2m, 2M, 2(m+M)$, though we will not explicitly use these moments in the analysis below.)

UPDATE, September 2019: Due to renewed interest in this question, I will expand upon my 2014 comment regarding why the above analysis appears at first glance to also lead to the incorrect conclusion that the disk rotation is also unstable if it one instead rotates around the $x$-axis (so that it is now the $M$-masses that are stationary and the $m$-masses that are rotating), or equivalently if one swaps the location of the $m$-masses and $M$-masses (which we will not do here to try to reduce confusion).

The reason for this is that centrifugal force $F_{\mathrm{Cent}} = -m \Omega \times \Omega \times r$ is only one of two inertial forces that is introduced when one is in a steadily rotating frame. The other inertial force introduced is the Coriolis force $F_{\mathrm{Cor}} = -2 m \Omega \times v$, which acts on moving bodies in the rotating reference frame in a direction orthogonal to the motion (and to the axis of rotation). Strictly speaking, one has to take into account the effect of both inertial forces when performing Newtonian mechanics in a steadily rotating frame. As it turns out, the Coriolis force has a negligible impact on the dynamics when rotating around the $y$-axis, but dominates the dynamics when rotating around the $x$-axis, which is why the preceding discussion is accurate in the former case but not the latter.

In contrast, suppose that the disk is close to the stable equilibrium state when it rotates around the $x$-axis. Working in a rotating frame around this axis, the $M$-masses are near the $x$-axis of rotation, while the $m$-masses lie near the $y$-axis. As before, the $M$-masses experience centrifugal force and thus begin drifting slightly away from the $x$-axis. But then the Coriolis force on these masses kicks in, which is now proportional to the heavy mass $M$ rather than the light mass $m$. As such, the rigid tension forces connecting the $M$-masses to the much lighter $m$-masses offer very little resistance to the Coriolis force, and the motion of the $M$-mass begins to rotate, constantly experiencing Coriolis acceleration in a direction orthogonal to its motion. This effect tends to make the $M$-mass rotate in a tight circle, and basically neutralises the net effect of the centrifugal force by rotating any outward motion away from the $x$-axis back into inward motion. The end result is that in the rotating frame, the disk wobbles a bit around its equilibrium state, but does not dramatically depart from it, which is what one would expect from a stable equilbrium (imagine now a marble rolling in the vicinity of a valley).

In contrast, suppose that the disk is close to the stable equilibrium state when it rotates around the $x$-axis. Working in a rotating frame around this axis, the $M$-masses are near the $x$-axis of rotation, while the $m$-masses lie near the $y$-axis. As before, the $M$-masses experience centrifugal force and thus begin drifting slightly away from the $x$-axis. But then the Coriolis force on these masses kicks in, which is now proportional to the heavy mass $M$ rather than the light mass $m$. As such, the rigid tension forces connecting the $M$-masses to the much lighter $m$-masses offer very little resistance to the Coriolis force, and the motion of the $M$-mass begins to rotate, constantly experiencing Coriolis acceleration in a direction orthogonal to its motion. This effect tends to make the $M$-mass rotate in a tight circle, and basically neutralises the net effect of the centrifugal force by rotating any outward motion away from the $x$-axis back into inward motion. The end result is that in the rotating frame, the disk wobbles a bit around its equilibrium state, but does not dramatically depart from it, which is what one would expect from a stable equilbrium (imagine now a marble rolling frictionlessly around the trough of a valley).