A cylinder of mass M and radius r is mounted on a frictionless axle over a wall. 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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$\frac{mg}{M + 2 m}$

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

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

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

answer is D.

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Detailed Solution

for bucket mg-T=ma..(1)

for cylinder $τ = Iα, τ = Tr and I = \frac{{Mr}^{2}}{2}$ ...(2)

Since there is no slipping $rα = a$

from (1) and (2) we get $T = \frac{ma}{2}$

On using the value of T we get $a = \frac{2 mg}{M + 2 m}$

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