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A uniform solid cylinder of mass M and radius R rotates about a frictionless axle. Two masses are suspended from a rope wrapped around the cylinder. If the system is released from rest, the tension in each rope is

\[\begin {array} {1 1} (a)\;\frac{Mmg}{M+m} & \quad (b)\;\frac{Mmg}{M+2m} \\ (c)\;\frac{Mmg}{M+3m} & \quad  (d)\;\frac{Mmg}{M+4m} \end {array}\]

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$mg-T=ma$. for each man.
$2TR= I \alpha$
$\qquad= \large\frac{MR^2}{2}. \alpha$
$T= \large\frac{MR \alpha}{4}$
Since string does not stop,
$a= R \alpha$
$\therefore mg= ma +\large\frac{Ma}{4}$
$\therefore a= \large\frac{4\;mg}{4m+M}$
$T= \large\frac{M}{4} \frac{4mg}{4m+M}$
$T= \large\frac{Mmg}{(M+4m)}$
answered Nov 27, 2013 by meena.p
edited Jun 19, 2014 by lmohan717
 

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