The pulleys in figure are identical, each having a radius R and moment of inertia I. Find the acceleration of the block M. 
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According to the question
$\text{Mg}-\text{T}_1=\text{Ma}\ \dots(1)$
$(\text{T}_1-\text{T}_2)\text{R}=\frac{\text{la}}{\text{R}}$
$\Rightarrow(\text{T}_2-\text{T}_1)=\frac{\text{la}}{\text{R}^2}\ \dots(2)$
$(\text{T}_2-\text{T}_3)\text{R}=\frac{\text{la}}{\text{R}^2}\ \dots(3)$
$\Rightarrow\text{T}_3-\text{mg}=\text{ma}\ \dots(4)$ By adding equation (2) and (3) we will get, $\Rightarrow(\text{T}_1-\text{T}_3)=2\frac{\text{la}}{\text{R}^2}\ \dots(5)$ By adding equation (1) and (4) we will get, $-\text{mg}+\text{Mg}+(\text{T}_3-\text{T}_1)=\text{Ma}+\text{ma}\ \dots(6)$ Substituting the value for $T_3 - T_1$ we will get $\Rightarrow\text{Mg}-\text{mg}=\text{Ma}+\text{ma}+\frac{2\text{la}}{\text{R}^2}$
$\Rightarrow\text{a}=\frac{(\text{M}-\text{m})\text{G}}{\big(\text{M}+\text{m}+\frac{\text{2l}}{\text{R}^2}\big)}$
$\text{Mg}-\text{T}_1=\text{Ma}\ \dots(1)$
$(\text{T}_1-\text{T}_2)\text{R}=\frac{\text{la}}{\text{R}}$
$\Rightarrow(\text{T}_2-\text{T}_1)=\frac{\text{la}}{\text{R}^2}\ \dots(2)$
$(\text{T}_2-\text{T}_3)\text{R}=\frac{\text{la}}{\text{R}^2}\ \dots(3)$
$\Rightarrow\text{T}_3-\text{mg}=\text{ma}\ \dots(4)$ By adding equation (2) and (3) we will get, $\Rightarrow(\text{T}_1-\text{T}_3)=2\frac{\text{la}}{\text{R}^2}\ \dots(5)$ By adding equation (1) and (4) we will get, $-\text{mg}+\text{Mg}+(\text{T}_3-\text{T}_1)=\text{Ma}+\text{ma}\ \dots(6)$ Substituting the value for $T_3 - T_1$ we will get $\Rightarrow\text{Mg}-\text{mg}=\text{Ma}+\text{ma}+\frac{2\text{la}}{\text{R}^2}$
$\Rightarrow\text{a}=\frac{(\text{M}-\text{m})\text{G}}{\big(\text{M}+\text{m}+\frac{\text{2l}}{\text{R}^2}\big)}$
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