The pulley shown in figure has a radius $10cm$ and moment of inertia $0.5kg-m^2$ about its axis. Assuming the inclined planes to be frictionless, calculate the acceleration of the $4.0kg$ block.
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$\text{m}_1\text{g}\sin\theta-\text{T}_1=\text{m}_1\text{a}\ \dots(1)$
$(\text{T}_1-\text{T}_2)=\frac{\text{la}}{\text{r}^2}\ \dots(2)$
$\text{T}_2-\text{m}_2\text{g}\sin\theta=\text{m}_2\text{a}\ \dots(3)$
Adding the equations (1) and (3) we will get
$\text{m}_1\text{g}\sin\theta+(\text{T}_2-\text{T}_1)-\text{m}_2\text{g}\sin\theta=(\text{m}_1+\text{m}_2)\text{a}$
$\Rightarrow(\text{m}_1-\text{m}_2)\text{g}\sin\theta=\Big(\text{m}_1+\text{m}_2+\frac{1}{\text{r}^2}\Big)\text{a}$
$\Rightarrow\text{a}=\frac{(\text{m}_1-\text{m}_2)\text{g}\sin\theta}{\Big(\text{m}_1+\text{m}_2+\frac{1}{\text{r}^2}\Big)}=0.248=0.25\text{ms}^{-2}$
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