Tidal Locking question about the Earth-Moon system

[Swamp Thing]

TL;DR Summary

Why should the moon's tidal bulge lag from the earth-moon axis if the rotation rate is slower than synchronous?

This section of the Wikipedia article on tidal locking https://en.wikipedia.org/wiki/Tidal_locking#Mechanism
states:
(Note : A is the earth; B is the moon)

The material of B exerts resistance to this periodic reshaping caused by the tidal force. In effect, some time is required to reshape B to the gravitational equilibrium shape, by which time the forming bulges have already been carried some distance away from the A–B axis by B's rotation. Seen from a vantage point in space, the points of maximum bulge extension are displaced from the axis oriented toward A. If B's rotation period is shorter than its orbital period, the bulges are carried forward of the axis oriented toward A in the direction of rotation, whereas if B's rotation period is longer, the bulges instead lag behind.

I am having trouble understanding that last "whereas" bit (bold font). It seems to me that if we consider lower and lower rotation speeds (especially sub-synchronous speeds) then what should happen is that the delay caused by "resistance to reshaping" would only become more and more negligible, so that the bulge would align itself more and more closely with the earth-moon axis. Essentially, the bulges "lead angle" with respect to the earth-moon axis should asymptotically approach zero as the rotation slows down.

Is the article wrong on this point, or am I?

 

[Swamp Thing]

Okay, I think I get how I was wrong.

Dear Moderator: Please delete the question.

 

[MarkL]

Don't delete yet.
To study tidal locking (resonance), I wish to model the moon (ellipsoid) as a dumbbell of length, L.
What is a proper length, L, for the dumbbell?
Centroid of half-ellipsoid is 3/8a, half-sphere is 3/8r. A sphere (moon) would produce no torque, so these values seem incorrect.

 

[hutchphd]

I think the primary effect is due to actual density fluctuations (mascons) in the nearly circular moon. I believe the actual shape of the moon is pretty much the usual oblate spheroid slightly flattened at the poles. Post what you find out.

 

[MarkL]

Subtract the sphere from the ellipsoid - the dumbbell is the remainder.

The equation for the ellipse is $x^2/a^2 + y^2/b^2 = 1$, $a>b$, and the equation of a sphere is $x^2 + y^2 = b^2$, of radius b. We find the centroid, $\overline{x}$, for half the volume as follows:
\begin{align}
\int{x dV} &= \pi \int{x y^{2}dx} = \int_{0}^{b} [\frac{b^{2}}{a^{2}} (xa^{2} - x^{3}) - (xb^{2} - x^{3})]dx\nonumber \\
&\qquad \qquad \qquad+ \int_{b}^{a}\frac{b^{2}}{a^{2}}(xa^{2} - x^{3})dx \nonumber \\
&\qquad \qquad\qquad\quad \quad= \pi \frac{b^{2}}{4}(a^{2}- b^{2}) \\
\int{dV} &= \frac{\pi}{2}(\frac{4}{3}ab^{2} - \frac{4}{3}b^{3}) \;= \pi \frac{2}{3}b^{2}(a - b) \\
\overline{x} &= \int{x dV} / \int{dV} \:\:= \frac{3}{8} (a + b) = \it{l}
\end{align}
Then, M$_{dumbbell}$ = (M - M$_{b}$)/2