Question

In the arrangement shown in the figure, two equal masses hung from light cords wrapped around a uniform solid cylinder of mass M and radius R. The cylinder is free to rotate about a horizontal axis. If the system is released from rest, then, the linear acceleration of each mass is ;

• A. $\frac{2mg}{M+4m}$
• B. $\frac{mg}{2M+m}$
• C. $\frac{4mg}{M+4m}$
• D. $\frac{mg}{M+4m}$

Answer 1

The correct option is C $\frac{4mg}{M+4m}$

$\begin{array}{c}\text{Moment of Inertia of Solid cylinder I}\\ I=\frac{M{R}^2}{2}\\ \text{Let the masses descent with an acceleration}\\ \text{(a) and the cylindur start rootating}\\ \text{with an angular accleration}\left(\alpha \right)\text{.}\\ \Rightarrow [a=R\alpha ]\text{The string do not slip.}\end{array}$

$\begin{array}{c}\Rightarrow \text{For constraint equation of mass}\\ mg-T=ma-\left(1\right)\\ \Rightarrow \text{For constraint equation of cylinder}\\ \begin{array}{c}\tau =I\alpha \tau =\text{torque}\\ 2T\cdot R=I\cdot \frac{a}{R}\Rightarrow T=\frac{M{R}^2}{2}\times \frac{a}{2R^2}=\frac{Ma}{4}-\left(2\right)\end{array}\\ a=\frac{4mg}{M+4m}\end{array}$