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(D) Let $a$ be the linear acceleration of the bucket and $T$ be the tension in the rope. For the bucket of mass $m$,the equation of motion is: $mg - T

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$\frac{2mg}{M + 2m}$

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(D) Let $a$ be the linear acceleration of the bucket and $T$ be the tension in the rope. For the bucket of mass $m$,the equation of motion is: $mg - T = ma$  --- $(1)$ For the cylinder of mass $M$ and radius $r$,the torque $	au$ is given by $	au = I\alpha$,where $I = \frac{1}{2}Mr^2$ is the moment of inertia of the cylinder and $\alpha = \frac{a}{r}$ is the angular acceleration. The torque is also given by $	au = Tr$. Equating the two expressions for torque: $Tr = (\frac{1}{2}Mr^2)(\frac{a}{r}) = \frac{1}{2}Mra$. Thus,$T = \frac{1}{2}Ma$ --- $(2)$ Substituting the value of $T$ from equation $(2)$ into equation $(1)$: $mg - \frac{1}{2}Ma = ma$ $mg = ma + \frac{1}{2}Ma = a(m + \frac{M}{2}) = a(\frac{2m + M}{2})$ $a = \frac{2mg}{M + 2m}$

$A$ cylinder of mass $M$ and radius $r$ is mounted on a frictionless axle over a well. $A$ rope of negligible mass is wrapped around the cylinder and a bucket of mass $m$ is suspended from the rope. The linear acceleration of the bucket will be

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