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

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Solution

Given in the question $: -$
Assume, Acceleration $= a$
angular acceleration, $\propto= \frac{a}{R}$
According to the question
$M g - T_{1} = M a . . . (1)$

$(T_{2} - T_{1}) R = I a / R =\Rightarrow (T_{2} - T_{1}) = I a / R^{2} . . . . (2)$

$(T_{2} - T_{3}) R = I / R^{2} . . . . (3)$
$\Rightarrow T_{3} - m g = m a . . . . (4)$

By adding equation $(2)$ and $(3)$ we will get.
$\Rightarrow (T_{1} - T_{3}) = 2 I a / R^{2} . . . . (5)$

By adding equation $(1)$ and $(4)$ we will get.
$- m g + M g + (T_{3} - T_{1}) = M a + m a$

Substituting the value for $T_{3} - T_{1}$ we will get
$\Rightarrow M g - m g = m a + m a + 2 I a / R^{2}$

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

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