Question

A uniform cylinder of mass $M$ and radius $R$ has a string wrapped around it. The string is held fixed and the cylinder falls vertically, as in figure. Show that the acceleration of the cylinder is downward with magnitude $a=\frac{2g}{3}$

Answer 1

$a=\frac{Mg-T}{M}$........(i)

$\alpha =\frac{TR}{\left(\frac{1}{2}M{R}^2\right)}=\frac{2T}{MR}$ ......(ii)
$a=R\alpha$.......(iii)
Solving these three equations, we get
$a=\frac{2g}{3}$
and $T=\frac{Mg}{3}$