Solve the following :


The pulleys in figure are identical each having a radius $\mathrm{R}$ and moment of inertia $\mathrm{I}$. Find the acceleration of block $M$.


$m g-T_{1}=m a-\cdots--(i)$

$\left(T_{1}-T_{2}\right) r_{1}=I \alpha$

$a=\alpha r$

$\left(T_{1}-T_{2}\right)=\frac{I \alpha}{R^{2}}$ For pulley $1-\cdots-($ ii $)$

$\left(T_{2}-T_{3}\right)=\frac{I \alpha}{R^{2}}$ For pulley $2 \cdots-\cdots($ iii $)$

For block of mass $m$

$T_{3}-m g=m a----(i v)$

$\mathrm{T}_{1}-T_{3}=\frac{2 I \alpha}{R^{2}}-\cdots-(i v)$

Adding equation (i) ans (iv)

$-m g+M g+\left(T_{3}-T_{1}\right)=M a+m a$

$M g-m g=M a+m a+\frac{I \alpha}{R^{2}}$

$\mathrm{Mg}-\mathrm{mg}=\mathrm{Ma}+\mathrm{ma}+\frac{1 \alpha}{R^{2}}$

$a=\frac{(M-m) g}{M a+m a+\frac{I \alpha}{R^{2}}}$

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