Unfolding Carpet

It could be easily solved by using conservation of energy. $U_i + K_i = U_f + K_I$

Note that the portion of the carpet that went flat will have no contribution to final potential energy if we take ground as reference line.

Also mass of the new cylinder will be $\frac{m}{4}$ $mgR= \frac{m}{4} g \frac{R}{2} + \frac{1}{2} \frac{m}{4} v^2 + \frac{1}{2} I \omega^2$ Did you observe that this is the case of pure rolling so , we can say $v = \frac{R}{2} \omega$

And also by the basics of rotational motion we know that Moment of Inertia of a cylinder is $\frac{M h^2}{2}$ about its axis passing through center if the circular faces.

So here it is $\frac{ m( \frac{R}{2})^2}{2}$

Plugging in the values and doing some more calculations will take you to $\boxed{v =\sqrt{\frac{14gR}{3}}}$