The acceleration of the cylinder in the downward direction is determined as follows:From the figure,
\( mg-2T=ma \)______ (1)
\( 2Tr=\frac{1}{2}m{r}^{2}\left(\frac{a}{r}\right)\)
\( 2T=\frac{1}{2}ma\) _______ (2)Substituting eqn. (2) in (1), we get
\( mg=ma+\frac{1}{2}ma\)
\( mg=\frac{3}{2}ma\)
\( a=\left(\frac{2}{3}\right)g\)
Since the centre of the cylinder moving away from rest position, the velocity \( v\) after it falls from distance \( h\) is,
\( {v}^{2}=2\left(\frac{2}{3}g\right)h\)
\( v=\sqrt{\frac{4gh}{3}}\)