Why, conceptually, do limaçons $r=a+b\cos\theta$ have dimples when $|\frac{a}{b}|<2$? Using calculus, I can justify that limaçons—the polar graphs of $r=a+b\cos\theta$ for various nonzero real values of $a$ and $b$—are dimpled when $|\frac{a}{b}|<2$, but that doesn't seem to yield any conceptual reason why that should be the case.  The boundary value 2 seems too nice to not have a conceptual explanation, so is there one?

The calculus:  For this sort of limaçon, with cosine, there are always vertical tangents to the graph at $\theta=k\pi$, and the dimple is characterized by a pair of vertical tangents near one of those two locations, but equally spaced before and after it, whereas a non-dimpled limaçon only has those two vertical tangents.  Vertical tangents occur when $\frac{dy}{dx}$ is undefined; for polars, that means when $\frac{dy}{d\theta}$ is undefined (for our limaçon, never) or when $\frac{dx}{d\theta}=0$.  For our limaçon, $\frac{dx}{d\theta}=-\sin\theta(a+2b\cos\theta)$, so $\frac{dx}{d\theta}=0$ implies $\sin\theta=0$ ($\theta=k\pi$) or $a+2b\cos\theta=0$.  This latter case, which can be rewritten as $\cos\theta=-\frac{a}{2b}$, has no solutions when $|\frac{a}{2b}|>1$, a single solution that is already in the solutions from $\sin\theta=0$ when $|\frac{a}{2b}|=1$ (so, no additional vertical tangents and hence no dimple when $|\frac{a}{b}|\ge 2$), and two solutions when $|\frac{a}{2b}|<1$ (so two additional vertical tangents and hence a dimple when $|\frac{a}{b}|<2$).
 A: Somewhat related to Robert's and Rahul's answers: it might be more illustrative to consider the curvature of the limaçon. Without loss of generality, consider the case $b=1$ and positive $a$ (all other cases are related by scaling and reflection about the horizontal axis). The curvature function is
$$\frac{a^2+3a\cos\,\theta+2}{\sqrt{(a^2+2a\cos\,\theta+1)^3}}$$
For $a=2$ (your boundary), a plot of the curvature function shows that its only zero is at $\theta=\pi$, and that point is a tangency point:

(Remember that a point of zero curvature corresponds to your curve being "locally straight").
For $a > 2$ or for $a < 1$, the curvature function is entirely positive (no "straight sections"). The cardioid case, $a=1$, yields a curvature function that is singular at $\theta=\pi$ (corresponding to the cusp). Finally, for the case $1 < a < 2$, the curvature becomes zero at the points $\theta=\arccos\dfrac{-a^2-2}{3a}$ and $\theta=2\pi-\arccos\dfrac{-a^2-2}{3a}$, which exactly corresponds to the case where one starts to see dimples...
A similar analysis can be seen in Gibson's Elementary Geometry of Differential Curves: An Undergraduate Introduction.
A: To put it in a wider context: 
the planar parametric curve ${\bf p}= {\bf p(t)}$ with velocity vector ${\bf v}$ and acceleration ${\bf a}$ is "turning to the left" if $v_1 a_2 - v_2 a_1 > 0$ and "turning to the right" if $v_1 a_2 - v_2 a_1 < 0$. For a polar curve, taking $t = \theta$, ${\bf p}(t) = r(t) [\cos(t), \sin(t)]$, 
$$v_1 a_2 - v_2 a_1 = r(t)^2 + 2 r'(t)^2 - r(t) r''(t)$$
In the case of your limaçons, with $r(t) = a + b \cos(t)$, this turns out to be $2 b^2+3 a b \cos(t)+a^2$.  The minimum, when $\cos(t) = \pm 1$ (depending on sign of $ab$) is 
$2 b^2 - 3 |a b| + a^2 = (|a|-|b|)(|a| - 2 |b|)$.  For this to be negative, producing your "dimple", we need $|b| < |a| < 2 |b|$.  
A: An answer is already given in the paper "Why do certain limacons have a dimple, Teaching Mathematics and its Applications Advance Access published March 29, 2006".
A: Here is a geometric way of looking at it. For the purposes of this answer, I shall be reparametrizing the limaçon as
$$r=2a(1-h\cos\,\theta)$$
(I consider $h$ here to be positive; negative values of $h$ correspond to a congruent curve reflected about the vertical axis.) I have chosen this form because this allows for a convenient phrasing of one of the definitions of a limaçon:

The limaçon is the locus of a point attached to a circle of radius $a$ rolling around the outside of a fixed circle of radius $a$, with the tracing point at a distance $ah$ from the center of the rolling circle.

In other words, the limaçon is a special case of the epitrochoid.
Your border case corresponds to $h=\frac12$. Interpreted from the epitrochoid viewpoint, this means that the tracing point is halfway between the circumference and the center of the rolling circle:

For $0 < h < \frac12$, the tracing point is near the center of the rolling circle, and thus the limaçon can be expected to be convex (with the extreme $h=0$ case giving the circle whose radius is twice the original). For $h > \frac12$, the tracing point "juts out" a bit, and thus we see the dimples in the limaçon:

For the cardioid case ($h=1$) in particular, the cusp corresponds to where the tracing point coincides with the point of contact of the fixed and rolling circles:

while for $h > 1$, the self-intersection point is expected due to the tracing point jutting out of the rolling circle.

As seen in this book, there is more than one way to define the limaçon geometrically, and I suppose those other definitions will also shed light on the differences between dimpled and undimpled limaçons. I think that treating the limaçon as an epitrochoid is the easiest to understand, though.
A: Let a particle move along the limaçon at a constant rate of $\dot\theta = 1$, and consider on the instant when $\theta = \pi$ and $r = a - b$.
If it were moving at a constant radius, its inward acceleration would be $r\dot\theta^2 = a - b$. But since its radius is also changing, you get an additional outward acceleration of $\ddot r = b$. These cancel out when $a = 2b$.
A: The invariant description of what you are looking for is that in a family of smooth curves, the appearance and disappearance of a "dimple" happens at curves where a tangent line has contact of order 4 with (at least one point of one branch of) the curve.
This is a relatively high order of local calculation and might not be amenable to a simple geometric argument about rolling circles.  The acceleration and curvature calculations are equivalent (because the two quantities are proportional to each other), and show contact of order at least 3.  The reflection symmetry of the curves enhances this to order 4; an even analytic function vanishing to $O(t^3)$ is $O(t^4)$.  Or one checks the signs of the curvature near the critical point to show it is a minimum.  In all these approaches some algebra is necessary to get the answer.
