A solid cylinder of radius $R$ rolls without slipping with a center of mass speed $v_0 = \sqrt{gR/3}$ on a horizontal surface with a vertical edge, as shown in the figure. Here, $g$ is the acceleration due to the gravity. At the moment when the cylinder loses contact with the surface due to rotation around the corner, the speed of its center of mass is:

Step 1: Identify the initial parameters of the solid cylinder.

The moment of inertia of a solid cylinder of mass $m$ and radius $R$ about its center of mass is:

$$I_{\text{cm}} = \frac{1}{2}mR^2$$

The cylinder rolls without slipping with an initial center of mass speed $v_0 = \sqrt{\frac{gR}{3}}$. The initial angular velocity is:

$$\omega_0 = \frac{v_0}{R}$$

Step 2: Apply conservation of angular momentum about the corner $O$.

As the cylinder reaches the edge, it begins to pivot about the corner $O$. The angular momentum about $O$ just before pivoting is:

$$L_O = I_{\text{cm}}\omega_0 + mRv_0$$ $$L_O = \frac{1}{2}mRv_0 + mRv_0 = \frac{3}{2}mRv_0$$

Using the parallel axis theorem, the moment of inertia about the corner $O$ is:

$$I_O = I_{\text{cm}} + mR^2 = \frac{3}{2}mR^2$$

Let $\Omega$ be the angular velocity immediately after pivoting begins. By conservation of angular momentum about $O$:

$$I_O \Omega = L_O$$ $$\frac{3}{2}mR^2\Omega = \frac{3}{2}mRv_0$$ $$\Omega = \frac{v_0}{R}$$

Step 3: Apply conservation of mechanical energy during pivoting.

Let $\phi$ be the angle of rotation of the center of mass about the corner $O$ from the vertical. By conservation of mechanical energy:

$$\frac{1}{2}I_O\omega^2 + mgR\cos\phi = \frac{1}{2}I_O\Omega^2 + mgR$$

Substitute $I_O = \frac{3}{2}mR^2$ and $\Omega = \frac{v_0}{R}$:

$$\frac{3}{4}mR^2\omega^2 + mgR\cos\phi = \frac{3}{4}mv_0^2 + mgR$$

Multiply the entire equation by $\frac{4}{3m}$:

$$R^2\omega^2 + \frac{4}{3}gR\cos\phi = v_0^2 + \frac{4}{3}gR$$ $$R^2\omega^2 = v_0^2 + \frac{4}{3}gR(1-\cos\phi)$$

Step 4: Determine the condition for losing contact.

The radial equation of motion for the center of mass is:

$$mg\cos\phi - N = mR\omega^2$$

The cylinder loses contact with the surface when the normal force $N = 0$:

$$R\omega^2 = g\cos\phi$$ $$R^2\omega^2 = gR\cos\phi$$

Step 5: Solve for the angle $\phi$ at the moment of losing contact.

Substitute $R^2\omega^2 = gR\cos\phi$ and $v_0^2 = \frac{gR}{3}$ into the energy equation:

$$gR\cos\phi = \frac{gR}{3} + \frac{4}{3}gR(1-\cos\phi)$$

Divide both sides by $gR$:

$$\cos\phi = \frac{1}{3} + \frac{4}{3}(1-\cos\phi)$$ $$\cos\phi = \frac{5}{3} - \frac{4}{3}\cos\phi$$ $$\frac{7}{3}\cos\phi = \frac{5}{3}$$ $$\cos\phi = \frac{5}{7}$$

Step 6: Calculate the speed of the center of mass at this moment.

The speed of the center of mass is $v = R\omega$.

$$v^2 = R^2\omega^2 = gR\cos\phi$$

Substitute $\cos\phi = \frac{5}{7}$:

$$v^2 = \frac{5gR}{7}$$ $$v = \sqrt{\frac{5gR}{7}}$$

Answer: The speed of the center of mass at the moment it loses contact is $v = \sqrt{\dfrac{5gR}{7}}$.