A solid cylinder length is suspended symmetrically through two massless strings, as shown in the figure. The distance from the initial rest position, the cylinder should be unbinding the strings to achieve a speed of $4 \mathrm{m} {\mathrm{s}}^{-1}$, is _____$\mathrm{cm}$.

(take $g=10 \mathrm{m} {\mathrm{s}}^{-2}$)

The portion of the strings between the ceiling and the cylinder are at rest. Hence, the points of the cylinder where the strings leave it are at rest. The cylinder is thus rolling without slipping on the strings. 

Now, from energy conservation

$\mathrm{mgh}=\frac{1}{2}m{v}^2+\frac{1}{2}\mathrm{I}{\omega }^2$

where, $I$ is moment of inertia of cylinder and $\omega =\frac{v}{R}$ is angular velocity.

Then, we have

 $$\begin{gathered} mgh=\frac{1}{2}m{v}^2+\frac{1}{2}\frac{m{R}^2}{2}{\omega }^2 \\ \Rightarrow gh=\frac{1}{2}{v}^2+\frac{1}{2}\frac{R^2}{2}{\left(\frac{v}{R}\right)}^2 \end{gathered}$$

$$\begin{gathered} 10h=\frac{16}{2}+\frac{16}{4} \\ \Rightarrow h=1.2 \mathrm{m}=120 \mathrm{cm} \end{gathered}$$