Question

A horizontal solid cylinder (of mass $m$) is pivoted about its longitudinal axis. To the end of a thread wrapped on the cylinder a block (of mass $2m$) is attached, as shown. If the system is left free, acceleration of the block (in ${\text{ms}}^{-2}$) is
(string is massless and there is no slipping anywhere)

Answer 1

From figure,
For $2m$,
Using Newtons second law
$2mg-T=2ma$

For $m$,
As $\tau =I\alpha$
using $a=r\alpha$, as there is no slipping
$Tr=\frac{1}{2}m{r}^2\frac{a}{r}$
$\Rightarrow 2T=ma\therefore 2mg-\frac{ma}{2}=2ma$
$\Rightarrow 2g=\frac{a}{2}+2a=\frac{5a}{2}$
$\therefore a=\frac{4g}{5}=8{\text{ms}}^{-2}$