Solution

Consider the moment when the center of the top cylinder has moved down a distance $x_T$ and the center of the bottom cylinder has moved to the right a distance $x_B$. At this moment, the cylinders are moving with speeds $v_T$ and $v_B$, respectively. From the geometry:

$$x_T = 2R(1 - \cos \theta).$$

Energy conservation gives:

$$\frac{1}{2}m(v_B^2 + v_T^2) = mgx_T = 2mgR(1 - \cos \theta).$$

From the geometry of the system:

$$x_B^2 + (2R - x_T)^2 = (2R)^2.$$

Taking the derivative with respect to time yields:

$$x_B v_B - (2R - x_T)v_T = 0,$$

which implies:

$$v_T = \frac{v_B x_B}{2R - x_T} = v_B \tan \theta.$$

Substitute $v_T$ into the energy conservation equation:

$$v_B^2 = \frac{4gR(1 - \cos \theta)}{1 + \tan^2 \theta}.$$

Simplify using the trigonometric identity $1 + \tan^2 \theta = \sec^2 \theta$:

$$v_B^2 = 4gR(1 - \cos \theta) \cos^2 \theta.$$

Find the maximum speed by setting the derivative with respect to $\theta$ equal to zero. This occurs at:

$$\cos \theta_m = \frac{2}{3}.$$

Substitute $\cos \theta_m$ into the equation for $v_B$:

$$v_B = \sqrt{\frac{16}{27}gR}.$$

The maximum speed of the bottom cylinder is:

$$\boxed{\sqrt{\frac{16}{27}gR}}.$$