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A cylinder of mass M and radius R is fixed on a frictionless axle over a well. A rope of negotiable mass is wrapped around the cylinder . A bucket of mass m is suspended from it. The linear acceleration of the bucket will be

$\begin{array}{1 1} (a)\;\frac{mg}{M+m} & \quad (b)\;\frac{2mg}{M+m} \\ (c)\;\frac{2mg}{M+2m} & \quad (d)\;\frac{2 mg}{2M+m} \end{array}$

1 Answer

$mg-T= ma$
$Tr= \large\frac{Mr^2}{2}. \alpha$
$T= \large\frac{Mr \; \alpha}{2}$
$\quad= \large\frac{Ma}{2}$
$mg= \bigg(\large\frac{M}{2}$$ +m\bigg)a$
$a= \large\frac{2m}{M+2m}g$
answered Nov 27, 2013 by meena.p
edited Jun 19, 2014 by lmohan717

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