A uniform solid cylinder of mass M and radius | vedclass

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(B) Let $T$ be the tension in each rope and $a$ be the downward acceleration of each mass $m$.For each mass $m$,the equation of motion is: $mg - T = m

Vedclass · Accepted Answer

$\frac{4mg}{M + 4m}$

Vedclass · Answer

(B) Let $T$ be the tension in each rope and $a$ be the downward acceleration of each mass $m$.For each mass $m$,the equation of motion is: $mg - T = ma$ ... $(1)$For the cylinder,the total torque $	au$ due to the two ropes is $	au = 2TR$.The moment of inertia of the solid cylinder about its axis is $I = \frac{1}{2}MR^2$.Using $	au = I\alpha$,where $\alpha$ is the angular acceleration: $2TR = (\frac{1}{2}MR^2)\alpha$.Since $a = \alpha R$,we have $\alpha = \frac{a}{R}$.Substituting $\alpha$ into the torque equation: $2TR = \frac{1}{2}MR^2(\frac{a}{R}) = \frac{1}{2}MRa$.Thus,$2T = \frac{1}{2}Ma$,or $T = \frac{1}{4}Ma$ ... $(2)$Substitute $T$ from $(2)$ into $(1)$: $mg - \frac{1}{4}Ma = ma$.$mg = ma + \frac{1}{4}Ma = a(m + \frac{M}{4}) = a(\frac{4m + M}{4})$.Therefore,$a = \frac{4mg}{M + 4m}$.

$A$ uniform solid cylinder of mass $M$ and radius $R$ rotates about a frictionless horizontal axle. Two similar masses $m$ are suspended with the help of two ropes wrapped around the cylinder. If the system is released from rest,then the acceleration of each mass will be

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