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 $2 m$ ) is attached, as shown. If the system is left free, acceleration of the block (in ${ms}^{- 2}$ ) is
(string is massless and there is no slipping anywhere)

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Solution

From figure,
For $2 m$ ,
Using Newtons second law
$2 m g - T = 2 m a$

For $m$ ,
As $τ = I α$
using $a = r α$ , as there is no slipping
$T r = \frac{1}{2} m r^{2} \frac{a}{r}$
$\Rightarrow 2 T = m a ∴ 2 m g - \frac{m a}{2} = 2 m a$
$\Rightarrow 2 g = \frac{a}{2} + 2 a = \frac{5 a}{2}$
$∴ a = \frac{4 g}{5} = 8 {ms}^{- 2}$

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