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From a disc of mass M and radius R, a smaller disc of radius $\large\frac{R}{2}$ is cut out as shown. The moment of inertia of the remaining portion about an axis through the centre 0 and perpendicular to the plane of the disc is

\[\begin {array} {1 1} (a)\;\frac{17MR^2}{32} & \quad (b)\;\frac{13MR^2}{32} \\ (c)\;\frac{11}{32}MR^2 & \quad  (d)\;\frac{MR^2}{2} \end {array}\]

1 Answer

Moment of inertia of the cut disc portion about $0=I_1$
Moment of inertia of remaining portion about $0=I_2$
$M.I$ of the full disc about $O=I$
Mass of out disc $=\large\frac{M}{R^2}. \bigg(\large\frac{R}{2}\bigg)^2$
$\qquad= \large\frac{M}{4}$
$I_1$ about $O^{1}=\large\frac{1}{2} \bigg( \large\frac{M}{4} \bigg) \bigg( \frac{R}{2} \bigg)^2$
$\qquad = \large\frac{MR^2}{32}$
$I_1$ about $O=\large\frac{MR^2}{32}+\frac{M}{4} \bigg( \frac{R}{4} \bigg)$
$\qquad= \large\frac{MR^2}{32}+\frac{MR^2}{16}$
$\qquad= \large\frac{3 MR^2}{32}$
$I_2= I-I_1$
$\qquad= \large\frac{MR^2}{2} -\frac{3MR^2}{32}$
$\qquad= \large\frac{13 MR^2}{32}$
answered Nov 26, 2013 by meena.p
edited Jun 19, 2014 by lmohan717

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