Rolling of body cone depends on whether ellipsoid is prolate or oblate?

[Happiness]

From the last few sentences of the below attached paragraph, when the inertia ellipsoid is prolate, the body cone rolls outside the space cone; when it is oblate, the body cone rolls inside the space cone.

Whether the body cone rolls outside or inside the space cone should depend on whether the body cone is rotating clockwise or anti-clockwise. Why should it depend on whether the ellipsoid is prolate or oblate?

The same sentences wrote that the ellipsoid is prolate when the (principal) moment of inertial about the symmetry axis is less than that about the other two principal axes. I believe it should be more than, rather than less than.

Suppose the symmetry axis is along the $x_1$ axis, then the principal moment of inertial about the symmetry axis is $I_1$, according to (5.35) and (5.33). And from Fig 5.4, the semi-principal axis along $x_1$, the symmetry axis, is longer than those along $x_2$ and $x_3$. Thus $I_1$ should be more than $I_2$ and $I_3$. Isn't it?

 

[marcusl]

From the last few sentences of the below attached paragraph, when the inertia ellipsoid is prolate, the body cone rolls outside the space cone; when it is oblate, the body cone rolls inside the space cone.

Whether the body cone rolls outside or inside the space cone should depend on whether the body cone is rotating clockwise or anti-clockwise. Why should it depend on whether the ellipsoid is prolate or oblate?

It can't possibly depend on the direction of rotation. For the frictionless, force- and torque-free conditions considered here, the motion continues forever. One can take a video of the motion and run it backwards with all the laws of physics being satisfied perfectly--but now the sense of rotation appears reversed. Hence the direction of rotation is irrelevant.

The relation between body and space cones is relevant, however. It is essentially geometric. One way to see it is to draw, or construct out of paper, the cones for prolate and oblate spheroids.

The same sentences wrote that the ellipsoid is prolate when the (principal) moment of inertial about the symmetry axis is less than that about the other two principal axes. I believe it should be more than, rather than less than.

No, this is correct. The moment of inertia reflects, in a sense, how far away from the rotation axis to put ring of matter containing the same mass as the body, such that the angular momentum is unchanged. (That distance is actually called the radius of gyration.) For a prolate spheroid (cigar shape), the radius of gyration about the symmetry axis is small. For an oblate spheroid of the same mass (flying saucer), it is large. This verifies the definition you quoted.

 

[Happiness]

Why does the inertia ellipsoid roll without slipping?

According to the paragraph attached below, it is because the instantaneous axis of rotation is momentarily at rest (with respect to the invariable plane or equivalently to the space axes). In the next instant, the inertia ellipsoid does not slip either. So that means the instantaneous axis of rotation is momentarily at rest too. And we can similarly argue that the instantaneous axis of rotation is always at rest. But this is not true since the instantaneous axis of rotation precesses in general.

 

[marcusl]

it is because the instantaneous axis of rotation is momentarily at rest (with respect to the invariable plane or equivalently to the space axes).

No, it says it is at rest in the body coordinates not the space coordinates.

 

[Happiness]

No, it says it is at rest in the body coordinates not the space coordinates.

But if we slip the inertia ellipsoid along the herpolhode, we can still have the vector $\rho$ invariant in the inertia ellipsoid since the point of contact does not change and the relative orientation of this point with respect to the rest of the ellipsoid (and to the body axes) does not change too. So it seems like slipping allows the instantaneous axis of rotation to be at rest in the body coordinates.

 

[marcusl]

I think the point is this: If you are slipping, then the body is not rotating around ρ but around the body's symmetry axis. (This is most obvious for pure slippage, that is, no friction). ρ ceases to be instantaneously at rest.

 

[Happiness]

Why does the inertia ellipsoid roll without slipping?

According to the paragraph attached below, it is because the instantaneous axis of rotation is momentarily at rest (with respect to the invariable plane or equivalently to the space axes). In the next instant, the inertia ellipsoid does not slip either. So that means the instantaneous axis of rotation is momentarily at rest too. And we can similarly argue that the instantaneous axis of rotation is always at rest. But this is not true since the instantaneous axis of rotation precesses in general.

View attachment 99530

I figured out the mistake. It is true that the instantaneous axis of rotation is always momentarily at rest. But this does not contradict the fact that the instantaneous axis of rotation precesses in general. This is because the instantaneous axis of rotation at time $t=t_1$ is different from that at time $t=t_2$ in general.

As an analogy, the point of contact of a rolling sphere is always momentarily at rest. But this does not contradict the fact that the point of contact (for a particular instant) moves about in a circle. This is because the point of contact at time $t=t_1$ is different from that at time $t=t_2$ in general.